If int fracleft(sqrt1 + x^2 + xright)^10left(sqrt1 + x^2 - xright)^9mathrmdx = frac 1m left(left(sqrt 1 + x ^ 2 + xright) ^ n left(n sqrt 1 + x ^ 2 - xright)right) + C$\int \frac{\left(\sqrt{1 + x^2} + x\right)^{10}}{\left(\sqrt{1 + x^2} - x\right)^9}\mathrm{d}x = \frac {1}{m} \left(\left(\sqrt {1 + x ^ {2}} + x\right) ^ {n} \left(n \sqrt {1 + x ^ {2}} - x\right)\right) + C$ where C$C$ is the constant of integration and mathbfm,mathbfnin mathbfN$\mathbf{m},\mathbf{n}\in \mathbf{N}$, then mathfrakm + mathfrakn$\mathfrak{m} + \mathfrak{n}$ is equal to

### Core Logic Let's simplify the integrand by rationalizing the denominator term block. Notice that: left(sqrt1+x^2 - xright)left(sqrt1+x^2 + xright) = (1+x^2) - x^2 = 1$$\left(\sqrt{1+x^2} - x\right)\left(\sqrt{1+x^2} + x\right) = (1+x^2) - x^2 = 1$$ frac1sqrt1+x^2 - x = sqrt1+x^2 + x$$\frac{1}{\sqrt{1+x^2} - x} = \sqrt{1+x^2} + x$$ Substituting this back into the denominator expression column: I = int left(sqrt1+x^2 + xright)^10 cdot left(sqrt1+x^2 + xright)^9 dx = int left(sqrt1+x^2 + xright)^19 dx$$I = \int \left(\sqrt{1+x^2} + x\right)^{10} \cdot \left(\sqrt{1+x^2} + x\right)^9 dx = \int \left(\sqrt{1+x^2} + x\right)^{19} dx$$ ### Step 1: Implementing the Substitution Path Let t = sqrt1+x^2 + x$t = \sqrt{1+x^2} + x$. Then: dt = left( fracxsqrt1+x^2 + 1 right) dx = left( fracx + sqrt1+x^2sqrt1+x^2 right) dx = fractsqrt1+x^2 dx$$dt = \left( \frac{x}{\sqrt{1+x^2}} + 1 \right) dx = \left( \frac{x + \sqrt{1+x^2}}{\sqrt{1+x^2}} \right) dx = \frac{t}{\sqrt{1+x^2}} dx$$ dx = fracsqrt1+x^2t dt$$dx = \frac{\sqrt{1+x^2}}{t} dt$$ Since sqrt1+x^2 + x = t$\sqrt{1+x^2} + x = t$ and sqrt1+x^2 - x = frac1t$\sqrt{1+x^2} - x = \frac{1}{t}$, adding both gives: 2sqrt1+x^2 = t + frac1t implies sqrt1+x^2 = frac12left(t + frac1tright)$$2\sqrt{1+x^2} = t + \frac{1}{t} \implies \sqrt{1+x^2} = \frac{1}{2}\left(t + \frac{1}{t}\right)$$ Thus, dx = frac12tleft(t + frac1tright) dt = frac12left(1 + frac1t^2right) dt$dx = \frac{1}{2t}\left(t + \frac{1}{t}\right) dt = \frac{1}{2}\left(1 + \frac{1}{t^2}\right) dt$. ### Step 2: Integrating with respect to t Substitute these back into the integral: I = int t^19 cdot frac12left(1 + frac1t^2right) dt = frac12 int left(t^19 + t^17right) dt$$I = \int t^{19} \cdot \frac{1}{2}\left(1 + \frac{1}{t^2}\right) dt = \frac{1}{2} \int \left(t^{19} + t^{17}\right) dt$$ I = frac12 left( fract^2020 + fract^1818 right) + C = fract^184 left( fract^210 + frac19 right) + C = fract^18360 big(9t^2 + 10big) + C$$I = \frac{1}{2} \left( \frac{t^{20}}{20} + \frac{t^{18}}{18} \right) + C = \frac{t^{18}}{4} \left( \frac{t^2}{10} + \frac{1}{9} \right) + C = \frac{t^{18}}{360} \big(9t^2 + 10\big) + C$$ ### Step 3: Matching Form and Finding m + n To match the template format, let's pull out a factor of t$t$: I = fract^19360 left( 9t + frac10t right) + C = fract^19360 left( 9left(sqrt1+x^2+xright) + 10left(sqrt1+x^2-xright) right) + C$$I = \frac{t^{19}}{360} \left( 9t + \frac{10}{t} \right) + C = \frac{t^{19}}{360} \left( 9\left(\sqrt{1+x^2}+x\right) + 10\left(\sqrt{1+x^2}-x\right) \right) + C$$ I = fracleft(sqrt1+x^2+xright)^19360 left( 19sqrt1+x^2 - x right) + C$$I = \frac{\left(\sqrt{1+x^2}+x\right)^{19}}{360} \left( 19\sqrt{1+x^2} - x \right) + C$$ Comparing this directly with the given answer format, we identify: - m = 360$m = 360$ - n = 19$n = 19$ Computing m + n$m + n$: m + n = 360 + 19 = 379$$m + n = 360 + 19 = 379$$ ### Pattern Recognition Expressions containing conjugate factors like sqrt1+x^2 pm x$\sqrt{1+x^2} \pm x$ frequently simplify under rationalization because their product equals 1. This dynamic quickly reduces fractional components into single power blocks. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Indefinite Integrals